If $f(x)=int_{3}^x sqrt{1+t^3} dt$ Find $(f^{-1})'(0)$
$begingroup$
Consider the integral $$f(x)=int_{3}^x sqrt{1+t^3} dt$$
Using Theorem 7 from my textbook which is $$frac{1}{f'(f^{-1}(a))}$$
$f'(t)= frac{2t^3}{2sqrt{1+t^3}}$
$f^{-1}(t)=sqrt[3]{t^2-1}$
I get confused here because does the integral affect how I should proceed or should I have incorporated the integral from the start?
calculus inverse-function
asked Dec 10 '18 at 23:39
Eric BrownEric Brown
737
$endgroup$
|
show 3 more comments
$begingroup$
Consider the integral $$f(x)=int_{3}^x sqrt{1+t^3} dt$$
Using Theorem 7 from my textbook which is $$frac{1}{f'(f^{-1}(a))}$$
$f'(t)= frac{2t^3}{2sqrt{1+t^3}}$
$f^{-1}(t)=sqrt[3]{t^2-1}$
I get confused here because does the integral affect how I should proceed or should I have incorporated the integral from the start?
calculus inverse-function
asked Dec 10 '18 at 23:39
Eric BrownEric Brown
737
$endgroup$
$begingroup$
Your derivative step is very wrong; a big red flag is the fact that you have a function of $x$ equal to a function of $t$.
$endgroup$
– T. Bongers
Dec 10 '18 at 23:49
$begingroup$
It was a force of habit, i'm used to $f(x)$
$endgroup$
– Eric Brown
Dec 10 '18 at 23:51
$begingroup$
It's still quite wrong. You've differentiated $sqrt{1 + t^3}$, not $f$. You've also just changed where the notation is incorrect to the previous line!
$endgroup$
– T. Bongers
Dec 10 '18 at 23:52
$begingroup$
Yeah so people like you don't comment on my tiny mistakes. I'm sleep deprived and over stressed, give me a break.
$endgroup$
– Eric Brown
Dec 10 '18 at 23:55
1
$begingroup$
The very first expression after “Consider the integral” is wrong: $f(t)$ should be $f(x)$. Also (here and in real life) I highly recommend that you don’t be rude to someone who’s trying to help you.
$endgroup$
– Shalop
Dec 11 '18 at 0:13
|
show 3 more comments
$begingroup$
Consider the integral $$f(x)=int_{3}^x sqrt{1+t^3} dt$$
Using Theorem 7 from my textbook which is $$frac{1}{f'(f^{-1}(a))}$$
$f'(t)= frac{2t^3}{2sqrt{1+t^3}}$
$f^{-1}(t)=sqrt[3]{t^2-1}$
I get confused here because does the integral affect how I should proceed or should I have incorporated the integral from the start?
calculus inverse-function
asked Dec 10 '18 at 23:39
Eric BrownEric Brown
737
$endgroup$
Consider the integral $$f(x)=int_{3}^x sqrt{1+t^3} dt$$
Using Theorem 7 from my textbook which is $$frac{1}{f'(f^{-1}(a))}$$
$f'(t)= frac{2t^3}{2sqrt{1+t^3}}$
$f^{-1}(t)=sqrt[3]{t^2-1}$
I get confused here because does the integral affect how I should proceed or should I have incorporated the integral from the start?
calculus inverse-function
calculus inverse-function
asked Dec 10 '18 at 23:39
Eric BrownEric Brown
737
asked Dec 10 '18 at 23:39
Eric BrownEric Brown
737
edited Dec 11 '18 at 0:15
Eric Brown
asked Dec 10 '18 at 23:39
Eric BrownEric Brown
737
asked Dec 10 '18 at 23:39
Eric BrownEric Brown
737
asked Dec 10 '18 at 23:39
Eric BrownEric Brown
737
737
$begingroup$
Your derivative step is very wrong; a big red flag is the fact that you have a function of $x$ equal to a function of $t$.
$endgroup$
– T. Bongers
Dec 10 '18 at 23:49
$begingroup$
It was a force of habit, i'm used to $f(x)$
$endgroup$
– Eric Brown
Dec 10 '18 at 23:51
$begingroup$
It's still quite wrong. You've differentiated $sqrt{1 + t^3}$, not $f$. You've also just changed where the notation is incorrect to the previous line!
$endgroup$
– T. Bongers
Dec 10 '18 at 23:52
$begingroup$
Yeah so people like you don't comment on my tiny mistakes. I'm sleep deprived and over stressed, give me a break.
$endgroup$
– Eric Brown
Dec 10 '18 at 23:55
1
$begingroup$
The very first expression after “Consider the integral” is wrong: $f(t)$ should be $f(x)$. Also (here and in real life) I highly recommend that you don’t be rude to someone who’s trying to help you.
$endgroup$
– Shalop
Dec 11 '18 at 0:13
|
show 3 more comments
$begingroup$
Your derivative step is very wrong; a big red flag is the fact that you have a function of $x$ equal to a function of $t$.
$endgroup$
– T. Bongers
Dec 10 '18 at 23:49
$begingroup$
It was a force of habit, i'm used to $f(x)$
$endgroup$
– Eric Brown
Dec 10 '18 at 23:51
$begingroup$
It's still quite wrong. You've differentiated $sqrt{1 + t^3}$, not $f$. You've also just changed where the notation is incorrect to the previous line!
$endgroup$
– T. Bongers
Dec 10 '18 at 23:52
$begingroup$
Yeah so people like you don't comment on my tiny mistakes. I'm sleep deprived and over stressed, give me a break.
$endgroup$
– Eric Brown
Dec 10 '18 at 23:55
1
$begingroup$
The very first expression after “Consider the integral” is wrong: $f(t)$ should be $f(x)$. Also (here and in real life) I highly recommend that you don’t be rude to someone who’s trying to help you.
$endgroup$
– Shalop
Dec 11 '18 at 0:13
$begingroup$
Your derivative step is very wrong; a big red flag is the fact that you have a function of $x$ equal to a function of $t$.
$endgroup$
– T. Bongers
Dec 10 '18 at 23:49
$begingroup$
Your derivative step is very wrong; a big red flag is the fact that you have a function of $x$ equal to a function of $t$.
$endgroup$
– T. Bongers
Dec 10 '18 at 23:49
$begingroup$
It was a force of habit, i'm used to $f(x)$
$endgroup$
– Eric Brown
Dec 10 '18 at 23:51
$begingroup$
It was a force of habit, i'm used to $f(x)$
$endgroup$
– Eric Brown
Dec 10 '18 at 23:51
$begingroup$
It's still quite wrong. You've differentiated $sqrt{1 + t^3}$, not $f$. You've also just changed where the notation is incorrect to the previous line!
$endgroup$
– T. Bongers
Dec 10 '18 at 23:52
$begingroup$
It's still quite wrong. You've differentiated $sqrt{1 + t^3}$, not $f$. You've also just changed where the notation is incorrect to the previous line!
$endgroup$
– T. Bongers
Dec 10 '18 at 23:52
$begingroup$
Yeah so people like you don't comment on my tiny mistakes. I'm sleep deprived and over stressed, give me a break.
$endgroup$
– Eric Brown
Dec 10 '18 at 23:55
$begingroup$
Yeah so people like you don't comment on my tiny mistakes. I'm sleep deprived and over stressed, give me a break.
$endgroup$
– Eric Brown
Dec 10 '18 at 23:55
1
1
$begingroup$
The very first expression after “Consider the integral” is wrong: $f(t)$ should be $f(x)$. Also (here and in real life) I highly recommend that you don’t be rude to someone who’s trying to help you.
$endgroup$
– Shalop
Dec 11 '18 at 0:13
$begingroup$
The very first expression after “Consider the integral” is wrong: $f(t)$ should be $f(x)$. Also (here and in real life) I highly recommend that you don’t be rude to someone who’s trying to help you.
$endgroup$
– Shalop
Dec 11 '18 at 0:13
|
show 3 more comments
1 Answer
1
active
oldest
votes
$begingroup$
HINTS:
From the Fundamental Theorem of Calculus, we have
$$f'(x)=sqrt{1+x^3}$$
Moreover, note that
$$f(3)=0implies f^{-1}(0)=3$$
Finally, the theorem from your text book reveals that
$$frac{df^{-1}(x)}{dx}=frac1{sqrt{1+(f^{-1}(x))^3}}$$
answered Dec 10 '18 at 23:53
Mark ViolaMark Viola
131k1275171
$endgroup$
$begingroup$
Is my inverse function correct though?
$endgroup$
– Eric Brown
Dec 11 '18 at 0:04
$begingroup$
Hi Eric. No, the inverse of $int_3^xsqrt{1+t^3},dt$ is not $sqrt[3]{x^2-1}$ .
$endgroup$
– Mark Viola
Dec 11 '18 at 0:11
add a comment |
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1 Answer
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active
oldest
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1 Answer
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active
oldest
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$begingroup$
HINTS:
From the Fundamental Theorem of Calculus, we have
$$f'(x)=sqrt{1+x^3}$$
Moreover, note that
$$f(3)=0implies f^{-1}(0)=3$$
Finally, the theorem from your text book reveals that
$$frac{df^{-1}(x)}{dx}=frac1{sqrt{1+(f^{-1}(x))^3}}$$
answered Dec 10 '18 at 23:53
Mark ViolaMark Viola
131k1275171
$endgroup$
$begingroup$
Is my inverse function correct though?
$endgroup$
– Eric Brown
Dec 11 '18 at 0:04
$begingroup$
Hi Eric. No, the inverse of $int_3^xsqrt{1+t^3},dt$ is not $sqrt[3]{x^2-1}$ .
$endgroup$
– Mark Viola
Dec 11 '18 at 0:11
add a comment |
$begingroup$
HINTS:
From the Fundamental Theorem of Calculus, we have
$$f'(x)=sqrt{1+x^3}$$
Moreover, note that
$$f(3)=0implies f^{-1}(0)=3$$
Finally, the theorem from your text book reveals that
$$frac{df^{-1}(x)}{dx}=frac1{sqrt{1+(f^{-1}(x))^3}}$$
answered Dec 10 '18 at 23:53
Mark ViolaMark Viola
131k1275171
$endgroup$
$begingroup$
Is my inverse function correct though?
$endgroup$
– Eric Brown
Dec 11 '18 at 0:04
$begingroup$
Hi Eric. No, the inverse of $int_3^xsqrt{1+t^3},dt$ is not $sqrt[3]{x^2-1}$ .
$endgroup$
– Mark Viola
Dec 11 '18 at 0:11
add a comment |
$begingroup$
HINTS:
From the Fundamental Theorem of Calculus, we have
$$f'(x)=sqrt{1+x^3}$$
Moreover, note that
$$f(3)=0implies f^{-1}(0)=3$$
Finally, the theorem from your text book reveals that
$$frac{df^{-1}(x)}{dx}=frac1{sqrt{1+(f^{-1}(x))^3}}$$
answered Dec 10 '18 at 23:53
Mark ViolaMark Viola
131k1275171
$endgroup$
HINTS:
From the Fundamental Theorem of Calculus, we have
$$f'(x)=sqrt{1+x^3}$$
Moreover, note that
$$f(3)=0implies f^{-1}(0)=3$$
Finally, the theorem from your text book reveals that
$$frac{df^{-1}(x)}{dx}=frac1{sqrt{1+(f^{-1}(x))^3}}$$
answered Dec 10 '18 at 23:53
Mark ViolaMark Viola
131k1275171
answered Dec 10 '18 at 23:53
Mark ViolaMark Viola
131k1275171
answered Dec 10 '18 at 23:53
Mark ViolaMark Viola
131k1275171
answered Dec 10 '18 at 23:53
Mark ViolaMark Viola
131k1275171
131k1275171
$begingroup$
Is my inverse function correct though?
$endgroup$
– Eric Brown
Dec 11 '18 at 0:04
$begingroup$
Hi Eric. No, the inverse of $int_3^xsqrt{1+t^3},dt$ is not $sqrt[3]{x^2-1}$ .
$endgroup$
– Mark Viola
Dec 11 '18 at 0:11
add a comment |
$begingroup$
Is my inverse function correct though?
$endgroup$
– Eric Brown
Dec 11 '18 at 0:04
$begingroup$
Hi Eric. No, the inverse of $int_3^xsqrt{1+t^3},dt$ is not $sqrt[3]{x^2-1}$ .
$endgroup$
– Mark Viola
Dec 11 '18 at 0:11
$begingroup$
Is my inverse function correct though?
$endgroup$
– Eric Brown
Dec 11 '18 at 0:04
$begingroup$
Is my inverse function correct though?
$endgroup$
– Eric Brown
Dec 11 '18 at 0:04
$begingroup$
Hi Eric. No, the inverse of $int_3^xsqrt{1+t^3},dt$ is not $sqrt[3]{x^2-1}$ .
$endgroup$
– Mark Viola
Dec 11 '18 at 0:11
$begingroup$
Hi Eric. No, the inverse of $int_3^xsqrt{1+t^3},dt$ is not $sqrt[3]{x^2-1}$ .
$endgroup$
– Mark Viola
Dec 11 '18 at 0:11
add a comment |
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$begingroup$
Your derivative step is very wrong; a big red flag is the fact that you have a function of $x$ equal to a function of $t$.
$endgroup$
– T. Bongers
Dec 10 '18 at 23:49
$begingroup$
It was a force of habit, i'm used to $f(x)$
$endgroup$
– Eric Brown
Dec 10 '18 at 23:51
$begingroup$
It's still quite wrong. You've differentiated $sqrt{1 + t^3}$, not $f$. You've also just changed where the notation is incorrect to the previous line!
$endgroup$
– T. Bongers
Dec 10 '18 at 23:52
$begingroup$
Yeah so people like you don't comment on my tiny mistakes. I'm sleep deprived and over stressed, give me a break.
$endgroup$
– Eric Brown
Dec 10 '18 at 23:55
1
$begingroup$
The very first expression after “Consider the integral” is wrong: $f(t)$ should be $f(x)$. Also (here and in real life) I highly recommend that you don’t be rude to someone who’s trying to help you.
$endgroup$
– Shalop
Dec 11 '18 at 0:13